Answer:
v_s = 34.269 m / s
Explanation:
This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.
f ’= f [tex]\frac{v}{v \mp v_s }[/tex]
where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer
In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together
5500 = f ([tex]\frac{v}{v - v_s }[/tex])
4500 = f ( [tex]\frac{v}{v + vs}[/tex])
let's solve these two equations
[tex]\frac{5500}{4500} = \frac{v+v_s}{v-v_s}[/tex]
1.222 (v-v_s) = v + v_s
v_s (1+ 1.22) = v (1.222 -1)
v_s = v [tex]\frac{0.222}{2.223}[/tex]
the speed of sound in air is v = 343 m / s
v_s = 343 0.09990
v_s = 34.269 m / s
